# On the equality of the rank-sum inequality

$\def \R{\mathbb{R}}$$\def \N{\mathbb{N}} \def \rank{\mbox{rank}} \def \range{\mbox{range}} Let m, n \in \N. Define M_{m, n}(\R) as a set of all m \times n matrices. For A, B \in M_{m, n}(\R) we have the following rank-sum inequality;$$ \rank(A + B) \leq \rank(A) + \rank(B).$$This follows from the observation that$\range(A+B) \subset \range(A) + \range(B)$. Here$\range(M) := \{ Mv : v \in \R^n \}$and$\rank(M) := \dim(\range(M))$for$M \in M_{m, n}(\R)$. The quesition is about when the equality holds in the above rank-sum inequality. According to Matrix Analysis by Roger A. Horn and Charles R.Johnson, the equality holds if and only if$\range(A) \cap \range(B) = \{0\}$and$\range(A^{T}) \cap \range(B^{T}) = \{0\}$. Here$A^T$denotes the transpose of$A$. I understand the "only if" part. Indeed suppse that the equlaity holds. Let$Av_1, \ldots , A v_k$be a basis of$\range(A)$and$Bw_1, \ldots, Bw_l$be a basis of$\range(B)$. Then$k = \rank{A}$,$l = \rank{B}$and the set$\{Av_1, \ldots, Av_k, Bw_1, \ldots, Bw_l\}$spans$\range(A+B)$. Since we are assuming that$\rank(A+B) = \rank(A) + \rank(B) = k + l$, the set$\{Av_1, \ldots, Av_k, Bw_1, \ldots, Bw_l\}$is independent. Therefore$\range(A) \cap \range(B) = \{0 \}$. Noting that$\rank(A) = \rank(A^T)$and etc, we have$\rank(A^T + B^T) = \rank(A^T) + \rank(B^T)$and thus$\range(A^T) \cap \range(B^T) = \{0\}$. However I don't know how to prove the "if" part. I tried to reduce to the case where$A$is of the form$I_k \bigoplus \large0$but that was not successful for me. Thank you for your help. • It seems to be a result of the subspace intersection Lemma :$ \dim(R(A) +R(B)) = \dim\; R(A) + \dim \; R(B) - \dim(R(A) \cap R(B)) $proved here on p.4 – C.C. Jun 5, 2022 at 1:43 ## 2 Answers Take a basis for the row space of$A$and a basis for the row space of$B$. Putting them together produces a linearly independent set, because the row spaces intersect in$\{0\}$only. Extend to a basis for$\mathbb R^m$. Multiplying an appropriate basis change matrix onto$A$and$B$from the right, you can achieve that$A$and$B$have their nonzero entries in different columns. Now, given any$v$in the column space of$A$and$w$in the column space of$B$, it is easy to find inputs$x,y$such that$(A+B)x=Ax=v$and$(A+B)y=By=w$. So the column space of$A+B$is the direct sum of the two column spaces. • Could you explain in more detail about "you can achieve that$A$and$B\$ have their nonzero entries in different columns" ?
– user438618
Jun 18, 2017 at 11:50

For the if part, we consider the singular value decomposition of $$A, B$$ as $$A = U_1 \Lambda_1 V_1^\top, B = U_2 \Lambda_2 V_2^\top$$ where $$U_1$$ is orthogonal $$m \times r_1$$, $$\Lambda_1$$ is diagonal $$r_1 \times r_1$$, $$V_1$$ is orthogonal $$n \times r_1$$, $$U_2$$ is orthogonal $$m \times r_2$$, $$\Lambda_2$$ is diagonal $$r_2 \times r_2$$, $$V_2$$ is orthogonal $$n \times r_2$$, for $$r_1 = rank(A), r_2 = rank(B)$$.

We notice that $$range(A) = range(U_1) ,\quad range(A^\top) = range(V_1) ,$$ $$range(B) = range(U_2) ,\quad range(B^\top) = range(V_2) .$$ The conditions $$range(A) \cap range(B) = \{0\}$$ and $$range(A^\top) \cap range(B^\top) = \{0\}$$ implies that $$U_1, U_2$$ are orthogonal and $$V_1, V_2$$ are orthogonal, so by extending them to $$m \times m$$ orthogonal matrix $$U = (U_1, U_2, U_3)$$ and $$n \times n$$ orthogonal matrix $$V = (V_1, V_2, V_3)$$, we can write $$A = (U_1, U_2, U_3) diag(\Lambda_1, 0, 0) (V_1, V_2, V_3)^\top ,\quad B = (U_1, U_2, U_3) diag(0, \Lambda_2, 0) (V_1, V_2, V_3)^\top .$$ It's easy to see from here that $$rank(A + B) = rank(A) + rank(B)$$.